Contributions to Algebra and Geometry Volume 43 (2002), No. 2, 303-324.

### Distance-preserving Maps in Generalized Polygons

### Part II: Maps on Points and/or Lines

Eline Govaert^{∗} Hendrik Van Maldeghem

*University of Ghent, Pure Mathematics and Computer Algebra*
*Galglaan 2, 9000 Gent, Belgium*

*e-mail: egovaert@@cage.rug.ac.be;* *hvm@@cage.rug.ac.be*

Abstract. In this paper, we characterize isomorphisms of generalized polygons (in particular automorphisms) by maps on points and/or lines which preserve a certain fixed distance. In Part I, we considered maps on flags. Exceptions give rise to interesting properties, which on their turn have some nice applications.

MSC 2000: 51E12

Keywords: generalized polygons, distance-preserving maps

1. Introduction

For an extensive introduction to the problem, we refer to Part I of this paper. Let us briefly describe the situation we are dealing with.

For the purpose of this paper, a generalized n-gon, n ≥3, is a thick geometry such that every two elements are contained in some ordinary n-gon, and no ordinary k-gons exist for k < n.

Given two generalized n-gons, and a map from either the point set or the point set and the line set of the first polygon to the point set, or the point set and the line set, respectively, of the second polygon, preserving the set of pairs of elements at a certain fixed distancei≤n, we investigate when this map can be extended to an isomorphism (“preserving” means that elements of the first polygon are at distanceiif and only if their images in the second polygon

∗The first author is a research assistant of the FWO, the*Fund for Scientific Research – Flanders (Belgium)*

0138-4821/93 $ 2.50 c 2002 Heldermann Verlag

are at distancei). In the first part of this paper, we dealt with the same problem considering
maps on the set of flags. More precisely, we showed (denoting the*order*of a polygon by (s, t)
if there are s+ 1 points on every line, and t+ 1 lines through every point):

Theorem 1. *Let* ∆ *and* ∆^{0} *be two generalized* m-gons, m≥2, let r *be an integer satisfying*
1≤ r ≤ m, and let α *be a surjective map from the set of flags of* ∆ *onto the set of flags of*

∆^{0}*. Furthermore, suppose that the orders of* ∆ *and* ∆^{0} *either both contain* 2, or both do not
*contain* 2. If for every two flags f, g *of* ∆, we have δ(f, g) = r *if and only if* δ(f^{α}, g^{α}) = r,
*then* α *extends to an (anti)isomorphism from* ∆ *to* ∆^{0}*, except possibly when* ∆ *and* ∆^{0} *are*
*both isomorphic to the unique generalized quadrangle of order* (2,2) *and* r= 3.

In [7], we also gave an explicit counterexample for the quadrangle of order (2,2).

In this part, we will show:

Theorem 2.

• *Let* Γ *and* Γ^{0} *be two generalized* n-gons, n ≥ 2, let i *be an even integer satisfying*
1≤i≤n−1, and let α *be a surjective map from the point set of* Γ *onto the point set*
*of* Γ^{0}*. Furthermore, suppose that the orders of* Γ *and* Γ^{0} *either both contain* 2, or both
*do not contain* 2. If for every two points a, b *of* Γ, we have δ(a, b) = i *if and only if*
δ(a^{α}, b^{α}) =i, then α *extends to an isomorphism from* Γ *to* Γ^{0}*.*

• *Let*Γ *and*Γ^{0} *be two generalized*n-gons,n ≥2, let i *be an odd integer satisfying*1≤i≤
n−1, and letα *be a surjective map from the point set of*Γ *onto the point set of* Γ^{0}*, and*
*from the line set of* Γ *onto the line set of*Γ^{0}*. Furthermore, suppose that the orders of* Γ
*and*Γ^{0} *either both contain*2, or both do not contain2. If for every point-line pair{a, b}

*of* Γ, we have δ(a, b) = i *if and only if* δ(a^{α}, b^{α}) = i, then α *extends to an isomorphism*
*from* Γ *to* Γ^{0}*.*

As already mentioned in Part I, there do exist counterexamples for the case n = i, and
we will construct some in Section 3, where we also prove two little applications. Also, the
condition i 6=n can be deleted for n = 3,4, of course, in a trivial way. For finite polygons,
the condition i 6= n is only necessary if n = 6 and the order (s, t) of Γ satisfies s = t. For
Moufang polygons, the condition i 6= n can be removed if Γ is not isomorphic to the split
Cayley hexagon *H(K) over some field* K (this is the hexagon related to the group G_{2}(K)).

We will prove these statements in Section 3.

To close this section, we repeat the notation that we use throughout the two parts of this
paper. Let Γ be a generalizedn-gon. For any point or linex, and any integeri≤n, we denote
by Γ_{i}(x) the set of elements of Γ at distance i from x, and we denote by Γ_{6=i}(x) the set of
elements of Γ not at distanceifromx. Ifκis a set of integers, then Γκ(x) is the set of elements
y of Γ satisfying δ(x, y) ∈ κ. If two elements are at distance n, then we say that they are
*opposite. Non-opposite elements*xandyhave a unique shortest chain (x=x_{0}, x_{1}, . . . , x_{k} =y)
of length k =δ(x, y) joining them. We denote that chain by [x, y], and we set x1 = proj_{x}y
(and hence x_{k−1} = proj_{y}x). When it suits us, we consider a chain as a set so that we can
take intersections of chains. For instance, if [x, z] = (x = x_{0}, x_{1}, . . . , x_{i}, x^{0}_{i+1}, . . . , x^{0}_{`} = z),
with no x^{0}_{j} equal to any xj^{0}, 0< i < j ≤k and i < j^{0} ≤ `, then [x, y]∩[x, z] = [x, xi]. If for
two non-opposite elements x, y the distance δ(x, y) is even, then there is a unique element

z at distance δ(x, y)/2 from both x and y; we denote z =x1y, or, if x and y are points at distance 2 from each other, then we also write xy:=x1y.

In the next section, we will prove Theorem 2. In Section 3, we produce some counterex- amples and prove some applications.

2. Proof of Theorem 2

As in the proof of Theorem 1, we again see that α is necessarily bijective.

We again prove the assertion in several steps, the general idea being to show that collinear- ity of points is preserved (then Lemma 1.3.14 of [9] gives the result).

Throughout we putT_{a,b}:= Γ_{i}(a)∩Γ_{i}(b), for points a, bof Γ.

2.1. Case i < ^{n−1}_{2} , with i even

This is the easy case. Putλ :={2i+ 2, . . . , n} 6=∅ and κ:={0,1, . . . , i−2}. Then one can
easily check that for two arbitrary distinct points a, b of Γ, we have T_{a,b} = ∅ if and only if
δ(a, b)∈ λ. Also, it is easily verified that, if δ(a, b) ∈/ λ∪ {i}, then Γ_{i}(a)∩Γ_{λ}(b) = ∅ if and
only if δ(a, b)∈κ. Now clearly, if δ(a, b)∈κ, then

Γ_{κ}(a)⊆Γ_{κ}(b)∪Γ_{i}(b)⇐⇒δ(a, b) = 2.

Hence α preserves collinearity and the assertion follows.

2.2. Case i = ^{n−1}_{2} , with i even

Here, Ta,b is never empty, for all pointsa, b of Γ.

First supposes >2. Let a, bbe arbitrary points of Γ. Then clearly |T_{a,b}|= 1 if and only
if δ(a, b) = 2i. So we can distinguish distance 2i. Also, it is clear that Γ_{i}(a)∩Γ_{2i}(b) = ∅ if
and only if δ(a, b)< i. Now one proceeds as in case 2.1.

Next suppose s = 2. Then |T_{a,b}| = 1 if and only if δ(a, b) ∈ {2i−2,2i}. Also, Γ_{i}(a)∩

Γ_{{2i−2,2i}}(b) =∅ if and only ifδ(a, b)< i−2. Then similarly as before, ifδ(a, b)< i−2, then

Γi(a)∩Γ<i−2(b) = ∅ ⇐⇒δ(a, b) = 2.

So again, α preserves collinearity.

The next two cases have some overlap. In particular, if i = (n+ 1)/2, then two proofs apply.

2.3. Case i ∈ {^{n}_{2} + 1,^{n+1}_{2} }, i even and n >6

LetSbe the set of pairs of distinct points (a, b) such thatδ(a, b)6=iand the setTa,b contains
at least two points at distance i from each other. We claim that a pair (a, b) belongs to S if
and only if δ(a, b)< i. Suppose first that 06=δ(a, b) =k, k < iand put m=a1b. Consider
a point c at distancei−k/2 from m such that proj_{m}a 6= w := proj_{m}c6= proj_{m}b (note that
δ(c, w) = i−k/2−1 > 0). Let v be the element of [c, w] at distance i/2 from c (such an
element exists since i/2≤i−k/2−1). Consider a pointc^{0} at distance i/2 from v such that

proj_{v}m 6= proj_{v}c^{0} 6= proj_{v}c. The points c and c^{0} are both points of T_{a,b} and lie at distance i
from each other.

Now let δ(a, b) = k > i and suppose by way of contradiction that c, c^{0} ∈ Ta,b with
δ(c, c^{0}) = i. If proj_{a}c6= proj_{a}c^{0}, then we have a path of length 2ibetweencandc^{0} containinga.

This implies that δ(c, c^{0})≥2n−2i > i, a contradiction. Suppose now that proj_{a}c= proj_{a}c^{0}.
Define v as [a, c] ∩[a, c^{0}] = [a, v]. If we put δ(a, v) = j, then there is a path of length

` = 2i−2j ≤ n between c and c^{0}. Now ` = i implies j = i/2. If the path [b, c] does not
contain v, there arises a circuit of length at most 3i < 2n, a contradiction (remembering
n > 6). But if v ∈ [b, c], there arises a path between a and b of length at most i, the final
contradiction. Our claim is proved.

Putκ={1,2, . . . , i−2}. Then (a, b)∈S if and only if δ(a, b)∈κ.

Now two distinct points a and b are collinear if and only if δ(a, b) ∈ κ and Γ_{κ}(a) ⊆
Γ_{κ}(b) ∪ Γ_{i}(b). Indeed, let δ(a, b) = k, 2 < k < i. Then k = i − j, 0 < j < i− 2.

Consider a point cat distance j+ 2 froma such that proj_{a}c6= proj_{a}b. Thenδ(a, c)∈κ, but
δ(b, c) = i+ 2. Ifδ(a, b) = 2, then the triangle inequality shows the assertion. Hence we can
distinguish distance 2 and so α preserves collinearity of points.

2.4. Case ^{n+1}_{2} ≤i < n−2, with i odd if i = ^{n}_{2} + 1

LetS be the set of pairs of points (a, b) for which there exists a point c, a6=c6=b, such that
T_{a,b} ⊆ Γ_{i}(c). Note that T_{a,b} is never empty because n/2 < i. We claim that the pair (a, b)
with δ(a, b) = k belongs to S if and only if 2 < k < 2(n−i). Note that there are always
even numbersk satisfying these inequalities becausei < n−2. So let a, bbe points of Γ with
δ(a, b) = k and let m be an element at distance k/2 from both a and b (m is unique if a is
not opposite b).

We first show that k < 2(n −i) if and only if every element y of T_{a,b} lies at distance
i−k/2 from m with proj_{m}a 6= proj_{m}y 6= proj_{m}b. Suppose first that k ≥ 2(n−i). Since
k+ 2i ≥ 2n, it is possible to find an element y of T_{a,b} such that proj_{a}y 6= proj_{a}b. Clearly,
δ(y, m) 6= i−k/2. Now suppose k < 2(n−i) and let y ∈ T_{a,b}. Let j be the length of the
path [a, b]∩[a, y]. Then there is a path of length ` =k−2j+i between b and y. If ` ≤n,
then ` = i and so necessarily j = k/2. Hence y is an element as claimed. If ` > n, then
δ(b, y)≥2n−` > i, a contradiction.

From this, it is clear that all pairs (a, b) with 2 < δ(a, b) = k < 2(n−i) belong to S
(indeed, choose the point c on the line proj_{a}b, at distance k from b).

We now show that

(∗) every point cat distanceifrom every element ofTa,bhas to lie at distancek/2 fromm,
and proj_{m}a= proj_{m}c or proj_{m}b= proj_{m}c.

Suppose cis a point at distance i from every element of T_{a,b}. Consider the set
T^{0} ={x∈T_{a,b}|δ(x, m) = i−k/2 and proj_{m}a6= proj_{m}x6= proj_{m}b}.

Then we may assume that T^{0} contains at least two elements yand y^{0} at distance 2 from each
other (this is clear either if i > n/2 + 1, or if n is odd, or if t > 2 in case i = n/2 + 1 and
i is odd; if t = 2 and i = n/2 + 1 with i odd, then we consider the dual of Γ; finally the

case i = n/2 + 1 with i even is not included in our assumptions, see Subsection 2.3). Put
w = y1y^{0}. Then δ(c, w) = i−1. Put γ = [c, w]. We show that γ contains m. Suppose by
way of contradiction that this is not true. Define the element z as [w, c]∩[w, m] = [w, z].

Putγ^{0} = [z, c]. An element y^{00} of T^{0} either lying onγ^{0} (if δ(c, m)≥i−k/2) or such that the
path [y^{00}, m] contains γ^{0} (otherwise), clearly does not lie at distanceifromc, a contradiction.

So the point c has to lie at distance k/2 from m. But if proj_{m}a 6= proj_{m}c 6= proj_{m}b, then
similarly we can find an element of T^{0} not at distance i fromc, which shows (∗).

Letk = 2 ork ≥2(n−i). We show that if a pointclies at distanceifrom every element of
Ta,b, thenc∈ {a, b}. Ifk= 2, then this follows immediately from (∗). So supposek≥2(n−i)
and let c be such a point. We may assume that, if δ(m, c) 6= n, then proj_{m}a = proj_{m}c. If
δ(a, c)6=n, then we define the element z as [m, c]∩[m, a] = [m, z]; otherwise we define z as
[proj_{m}a, c]∩[proj_{m}a, a] = [proj_{m}a, z]. Note that δ(c, z) =δ(a, z) =:`. Put j =i−n+k/2.

It is easy to check that for an elementv of the path [a,proj_{m}a], the following property holds.

(∗∗) There exists y∈T_{a,b} such that [a, y]∩[a, m] = [a, v] if and only if δ(a, v)≤j.

It follows from (∗∗) that`≤j (indeed, if y∈Ta,b is such thatm /∈[a, y], then the path [c, y]

is longer than the path [a, y]).

But similarly, if `≤j, then an elementy ∈T_{a,b} such that proj_{z}c∈ [a, y], does not lie at
distancei from c, a contradiction.

This shows our claim. Note that, if 2< k <2(n−i) and if c is a point of Γ at distance
k/2 from m with proj_{m}c∈ {proj_{m}a,proj_{m}b}, then automaticallyT_{a,b}⊆Γ_{i}(c).

Now defineS^{0} ={(a, c)| ∃b ∈ P such that Ta,b ⊆Γi(c)}. From the previous paragraph it
is clear that S^{0} \S is precisely the set of all pairs of collinear points.

Hence α preserves collinearity.

2.5. Case i = n−2 2.5.1. Case n = 6

Let C be the set of pairs of points (a, b), δ(a, b) 6= 4, such that for every point y in T_{a,b},
there exists a point y^{0} in T_{a,b}, y^{0} 6=y and δ(y, y^{0}) 6= 4, with the property that Γ_{4}(y)∩T_{a,b} =
Γ4(y^{0})∩Ta,b. Clearly, C contains all pairs of collinear points. Suppose now that (a, b) ∈ C
with δ(a, b) = 6. We look for a contradiction. If x is a point of T_{a,b}, then either x lies on a
line at distance 3 from both a and b, or x is a point at distance 3 from a line A through a
and from a line B through b, with A opposite B. Then one can check that for a point y of
T_{a,b} on a line at distance 3 from both aand b, there does not exist a pointy^{0} 6=y inT_{a,b} such
that Γ_{4}(y)∩T_{a,b} = Γ_{4}(y^{0})∩T_{a,b}. So the set C is the set of pairs of collinear points and the
theorem follows.

2.5.2. Case n >6 Step 1: the set Sa,b

For two pointsa and b, we define

S_{a,b} ={x∈ P |Γ_{n−2}(x)∩T_{a,b}=∅}.

We claim the following:

(i) If δ(a, b) = 2 ands ≥3, then S_{a,b} = (Γ_{2}(a)∪Γ_{2}(b))\Γ_{1}(ab). If δ(a, b) = 2 ands = 2,
then S_{a,b}= (Γ_{2}(a)∪Γ_{2}(b))\ {a, b}.

(ii) If δ(a, b) = 4, then {a1b} ⊆ S_{a,b} = {a1b} ∪[Γ_{2}(a1b)∩Γ_{4}(a)∩Γ_{4}(b)]. If t ≥ 3, then
S_{a,b} = {a1b}. If k := δ(a, b) 6∈ {2,4, n}, then every x ∈ S_{a,b} lies at distance k/2 from
a1b =: m with proj_{m}a 6= proj_{m}x 6= proj_{m}b. If moreover s > 2 and k ≡ 2 mod 4, or
t >2 andk ≡0 mod 4, then S_{a,b} =∅. Finally, ifδ(a, b) = n, then let γ be an arbitrary
path of lengthnjoiningaandb, letmbe the middle element ofγ and putv_{a}= proj_{m}a,
v_{b} = proj_{m}b. Then

S_{a,b} ⊆ (Γ_{n/2}(m)∩Γ_{n/2+1}(v_{a})∩Γ_{n/2+1}(v_{b}))
[

(Γ_{n/2+1}(v_{a})∩Γ_{n/2+2}(m)∩Γ_{n}(a))
[

(Γ_{n/2+1}(v_{b})∩Γ_{n/2+2}(m)∩Γ_{n}(b)).

If moreover s >2 and n≡2 mod 4, or t >2 and n≡0 mod 4, then
S_{a,b} ⊆ (Γ_{n/2+1}(v_{a})∩Γ_{n/2+2}(m)∩Γ_{n}(a))

[

(Γ_{n/2+1}(v_{b})∩Γ_{n/2+2}(m)∩Γ_{n}(b)).

We prove these claims.

(i) Suppose δ(a, b) = 2. Clearly, every point collinear with a or b, not on the line ab,
belongs toS_{a,b}. Also, if s= 2, then the unique point of abdifferent from a and b is an
element ofS_{a,b}. Let xbe an arbitrary point inS_{a,b}. Put j =δ(x, a). If j =s = 2, then
there is nothing to prove, so we may assume (j, s) 6= (2,2). Suppose first there exists
a j-path γ between a and x containing ab, but not the point b. Let v be the element
on γ at distance j/2 from a, and consider an element y at distance n−2−j/2 from
v such that proj_{v}a 6= proj_{v}y 6= proj_{v}x. Note that such an element v exists because
(j, s) 6= (2,2). Then y lies at distance n−2 from a, b and x, a contradiction. So we
can assume that proj_{ab}x = a. If j = 2, then again, there is nothing to prove. So we
may assume 2 < j < n (the case j = n is contained in the previous case, or can be
obtained from the present case by interchanging the roles of a and b). Let v be an
element at distance n−j −1 from the line ab such that a 6= proj_{ab}v 6= b. Note that
v and x are opposite and δ(a, v) = n −j. Consider an element v^{0} incident with v,
different from proj_{v}a, and let v^{00} be the projection of x onto v^{0}. Let w be the element
of [x, v^{00}] at distance j/2−2 from v^{00}. An element y at distance j/2−2 from w such
that proj_{w}x6= proj_{w}y6= proj_{w}v^{00} lies at distancen−2 froma,b and x, a contradiction.

Claim (i) is proved.

(ii) We proceed by induction on the distance k between a and b, the case k = 2 being
claim (i) above. Supposeδ(a, b) =k > 2 and let m be an element at distancek/2 from
botha and b. Note that, if δ(a, b) = 4, the point a1b indeed belongs toS_{a,b}. Let now
x be an arbitrary element of S_{a,b} and put `=δ(x, m).

Suppose first that, if ` 6= n, proj_{m}a 6= proj_{m}x 6= proj_{m}b. Then we have the following
possibilities :

1. Suppose ` < k/2. Then δ(a, x)< k and we apply the induction hypothesis. Since
b ∈ S_{a,x} 6= ∅ and m 6= a1x, we have δ(a, x) ∈ {2,4}. Hence either δ(a, b) = 4
and x = a1b (which is a possibility mentioned in (ii)), or δ(a, b) = 6 and x lies
onm, or δ(a, b) = 8 and x =m. But in these last two cases, the “position” of b
contradicts the induction hypothesis.

2. Suppose` ≥k/2. Letγ^{00} be an`-path betweenm andxcontaining neither proj_{m}a
nor proj_{m}b. Putγ^{0} = [a, m]∪γ^{00}. Letwbe the element onγ^{0} at distance (`+k/2)/2
from both x and a. If ` =k/2 (and hence w = m) and either k ≡ 2 mod 4 and
s = 2, or k ≡ 0 mod 4 and t = 2, then there is nothing to prove. Otherwise,
there exists an element y of Γ at distance n−2−(k/2 +`)/2 from w such that
proj_{w}a 6= proj_{w}y 6= proj_{w}x and proj_{w}b 6= proj_{w}y. Now y lies at distance n−2
froma, b and x, a contradiction.

Let nowxbe a point ofS_{a,b}at distance`fromm, 0< ` < n, for which proj_{m}x= proj_{m}a.

Let [a, m]∩[x, m] = [v, m], and put i^{0} =δ(v, a). We have the following possibilities :
1. Suppose `≤k/2 or `=k/2 + 2 and i^{0} < k/2−1. Againδ(a, x)< k and applying

the induction hypothesis, we obtain a contradiction as in Case 1 above.

2. Suppose n > ` > k/2 + 2. Leth be an element at distancen−2 fromxsuch that
proj_{m}a6= proj_{m}h6= proj_{m}b andδ(m, h) =n−`+ 2. Letj =n−2−δ(h, m)−k/2.

Leth^{0} be the element on the (n−2)-path between x and h at distance j/2 from
h. An elementy at distancej/2 fromh^{0} such that proj_{h}0x6= proj_{h}0y 6= proj_{h}0hlies
at distancen−2 from a, b and x, a contradiction.

3. Suppose ` =k/2 + 2, i^{0} =k/2−1 andk < n−1. Then δ(b, x) =k+ 2 andv lies
at distancek/2 + 1 from bothb andx. Let Σ be an apartment containingx,b and
v, and letv^{0} be the element in Σ opposite v. Letw= proj_{v}a, w^{0} = proj_{v}0wand d
the length of the path [w, a]∩[w, w^{0}]. Note thatd ≤k/2−2. For an elementynot
opposite w^{0}, let w_{y}^{00} be the element such that [w, w^{0}]∩[y, w^{0}] = [w^{00}_{y}, w^{0}]. Consider
now an elementy such that δ(w^{00}_{y}, w^{0}) =k/2−d−2 andδ(w^{00}_{y}, y) =d. Thenylies
at distancen−2 from a, b and x, a contradiction.

4. If `=k/2 + 2,i^{0} =n/2−1 and k =n, there is nothing to prove.

5. Suppose finally` =k/2 + 2,i^{0} =k/2−1 and k =n−1. Then δ(b, x) =n−1. Let
b^{0} and x^{0} be the elements of the path [b, x] at distance (n−1)/2−1 from b and
x, respectively. Since a ∈S_{b,x}, either δ(a, b^{0}) = (n+ 1)/2 or δ(a, x^{0}) = (n+ 1)/2
(this is what we proved up to now for the “position” of a point ofS_{b,x}). But since
we obtain a path betweena and b^{0} (x^{0}) of lengthd= (3n−5)/2 (passing through
proj_{m}a), the triangle inequality implies δ(a, b^{0}), δ(a, x^{0}) ≥ 2n−d > (n + 1)/2, a
contradiction.

This completes the proof of our claims.

Step 2: the sets O and O

Ifs≥3 andt ≥3, letO be the set of pairs of distinct points (a, b) such that|Sa,b|>1. Then O contains only pairs of collinear points and pairs of opposite points, and all pairs of collinear

points are included inO. Note that, if n is odd, there are no pairs of opposite points, which concludes the proof in this case.

Let now s= 2 or t= 2 (so n is even). For a point c∈Sa,b, we define the set
C_{a,b;c} ={c^{0} ∈S_{a,b}|S_{c,c}^{0} ∩ {a, b} 6=∅}.

Now let O be the set of pairs of points (a, b), δ(a, b) 6= n − 2 for which |S_{a,b}| > 1 and

|C_{a,b;c}| > 1, ∀c ∈ S_{a,b}. Note that if a and b are collinear, the pair (a, b) always belongs
toO. Clearly, no pair of points at mutual distance 4 belongs to O. Now consider two points
a and b at distance k, 4 < k < n −2. We show that such a pair (a, b) does not belong
to O. Putm =a1b and let x be a fixed point of S_{a,b}. Let x^{0} be an element of S_{a,b} different
from x. Since s = 2 or t = 2, we have proj_{m}x = proj_{m}x^{0}, so δ(x, x^{0}) ≤ k −2. But now
δ(a, x1x^{0}) = δ(b, x1x^{0})≥k/2 + 1> δ(x, x^{0})/2, so neitheranorb belongs toS_{x,x}^{0}. This shows
thatO contains only pairs of collinear or opposite points, and all pairs of collinear points are
included in O.

Let O be the set of pairs of points (a, b) satisfying δ(a, b) 6= n−2, (a, b) 6∈ O and such that there exist a point c∈Sa,b for which (a, c) and (b, c) both belong to O. Then clearly, O contains all pairs (a, b) of points at mutual distance 4 (indeed, consider the point c=a1b), and also some pairs of opposite points (possibly none, or all).

Step 3: the set O^{0}
Suppose first s≥3.

Let O^{0} be the subset of O of pairs (a, b) for which there exist points c and c^{0} such that the
following conditions hold :

(i) T_{a,b} ⊆Γ_{n−2}(c)∪Γ_{n−2}(c^{0}),T_{c,c}^{0} ⊆Γ_{n−2}(a)∪Γ_{n−2}(b);

(ii) (c, c^{0}),(a, c),(a, c^{0}),(b, c),(b, c^{0})∈O;

(iii) (c, y),(c^{0}, y)6∈O, ∀y∈Ta,b.

We claim that O^{0} is the set of pairs of collinear points. A pair (a, b) of collinear points
always belongs to O^{0}. Indeed, here, we can choose c and c^{0} on the line ab, different from
a and b (Condition (iii) is satisfied because n 6= 6). So let δ(a, b) = n and suppose by
way of contradiction that we have two points c and c^{0} with the above properties. Let m ∈
Γ_{n/2}(a)∩ Γ_{n/2}(b). For an element x at distance j from m, 0 ≤ j ≤ n/2− 3, such that
proj_{m}a6= proj_{m}x6= proj_{m}b, define the following set:

T_{x} ={y∈T_{a,b}|δ(x, y) =n/2−2−j,proj_{x}a6= proj_{x}y6= proj_{x}b}.

Note thatT_{x} ⊆T_{a,b}. We first prove that for any set T_{x},

(3) there does not exist a point v ∈ {c, c^{0}} such thatT_{x}⊆Γ_{n−2}(v).

Put M = Γn/2(a)∩Γn/2(b). Suppose Tm ⊆ Γn−2(v), with v ∈ {c, c^{0}}. It is easy to see (see
for instance Step 4 of Subsection 3.4 of [7]) that δ(v, m) = n/2 and proj_{m}a = proj_{m}v or
proj_{m}b= proj_{m}v. Suppose proj_{m}a= proj_{m}v. But then δ(a, v)≤n−2, so δ(a, v) = 2 (since
(a, v)∈O) andvis a point at distancen/2 frommlying on the lineL= proj_{a}m. This implies
that for an arbitrary point m^{0} of M, m^{0} 6=m, T_{m}^{0} ∩Γ_{n−2}(v) =∅ (note that T_{m}∩T_{m}^{0} =∅),

so T_{m}^{0} ⊆ Γ_{n−2}(v^{0}), with {v, v^{0}} = {c, c^{0}}. We obtain a contradiction by considering a third
element of M.

Letx be an element at distance j = 1 from m such that proj_{m}a6=x6= proj_{m}b. Suppose
T_{x} ⊆Γ_{n−2}(v), withv ∈ {c, c^{0}}. Then again it is easy to show that δ(v, x) = δ(x, a) = n/2 + 1
and proj_{x}v = proj_{x}a =m. If proj_{m}a= proj_{m}v or proj_{m}b = proj_{m}v, then we are back in the
previous case, which led to a contradiction, so suppose proj_{m}a 6= proj_{m}v 6= proj_{m}b. Consider
the n-path between aand v that containsm. Then we can find a pointyof T_{a,b} on this path
that is collinear with v, in contradiction with condition (iii). Note that thus no element of
{c, c^{0}} lies at distance n/2 from m.

We now proceed by induction on the distance j between x and m. Let j > 1. Consider
an element xat distancej fromm such that proj_{m}a6= proj_{m}x6= proj_{m}b. Suppose by way of
contradiction that Tx ⊆Γn−2(v), with v ∈ {c, c^{0}}. Letx^{0} = proj_{x}m. Then it is again easy to
show that δ(v, x) =δ(a, x) =n/2 +j and proj_{x}v =x^{0}. Remark that proj_{x}0a6= proj_{x}0v, since
otherwiseT_{x}^{0} ⊆Γ_{n−2}(v) (sinceδ(v, x^{0}) =n/2 +j−1 andδ(v, w) =n/2−j−1, withw∈T_{x}^{0}),
in contradiction with the induction hypothesis. Suppose first that in the case j = 2, t≥3 or
n≡2 mod 4. Consider now an elementz incident with the elementw= proj_{x}0a, but different
from proj_{w}a, from proj_{w}b and from x^{0} (such an element exists, because of the restrictions
above). But then we have δ(v, w^{0}) = n, for every element w^{0} of Tz, so Tz ⊆ Γn−2(v^{0}), with
{c, c^{0}}={v, v^{0}}, a contradiction with the induction hypothesis.

Now let j = 2, t = 2 and n ≡ 0 mod 4. Let L be the line mx and put w = proj_{L}v.

Then Tw ⊆Γn−2(v^{0}), with {v, v^{0}} ={c, c^{0}}, sov^{0} is a point at distance n/2 + 2 from m with
δ(m, v^{0}) 6= n/2, and proj_{L}v^{0} 6= w. Now consider the point on [m, b] at distance n/2−4
from m. This is a point of T_{c,c}^{0}, but it does not lie at distance n−2 from a, nor from b, in
contradiction with condition (i). This completes the proof of (3).

Consider now a lineLat distance j =n/2−3 from m, such that proj_{m}a6= proj_{m}L6= proj_{m}b.

The points onLdifferent from the projection ofm ontoLare points ofT_{L}. By (3), we know
that TL 6⊆ Γn−2(v), for v ∈ {c, c^{0}}. Since s ≥ 3, TL contains at least 3 points, so we may
suppose that at least two points of them are contained in Γ_{n−2}(v), with v ∈ {c, c^{0}}. This
implies that v is at distancen−3 fromL, so at distancen−4 from a unique pointxof L. If
x= proj_{L}a, thenTL ⊆Γn−2(v), a contradiction, so we can assume thatx6= proj_{L}a. Let first
n6= 8 or t≥3. Then consider a line L^{0} incident with proj_{L}a,L^{0} 6=L, at distance n−3 from
botha and b (such a line always exists because of our assumptions). NowT_{L}^{0} ∩Γ_{n−2}(v) = ∅
(because all points ofTL^{0} lie oppositev), soTL^{0} is contained in Γn−2(v^{0}), with{v, v^{0}}={c, c^{0}},
the final contradiction.

Let now n = 8 and t = 2. Then δ(v^{0}, x) = 6, with {v, v^{0}} = {c, c^{0}}, T_{L} 6⊆ Γ_{6}(v^{0}) and
δ(v, v^{0})∈ {2,8}. Now for each potential v^{0}, it is possible to construct a point of Tv,v^{0} not at
distancen−2 from a nor from b, a contradiction with condition (i). For example, let us do
in detail the case δ(v, v^{0}) = 2. Since v does not lie at distance 6 from a or b, we know that
δ(a, vv^{0}) = δ(b, vv^{0}) = 7, v^{0} 6= proj_{vv}0a 6= v and v^{0} 6= proj_{vv}0b 6= v. Also proj_{vv}0a 6= proj_{vv}0b,
since otherwise we would obtain a point ofT_{a,b} not at distance 6 fromv nor fromv^{0}. Now let
N be the line at distance 3 fromb and at distance 4 fromvv^{0}. Then the points ofN different
from proj_{N}v are points of Tv,v^{0} \Γ6(b), but not all these points lie at distance 6 from a, a
contradiction.

So in the caseδ(a, b) = n, pointscand c^{0} with the above properties cannot exist, and the

proof is finished for s≥3.

Suppose nows = 2.

Let first n = 8. Note that t ≥ 4. Let O^{0} be the subset of O of pairs of points (a, b) for
which there exist a pointc∈S_{a,b}and pointsc^{0},c^{00} belonging toC_{a,b;c} such thatS_{c,c}^{0}∩ {a, b}=
S_{c,c}^{00}∩ {a, b}=S_{c}^{0}_{,c}^{00}∩ {a, b}={a}. We claim thatO^{0} is the set of all pairs of collinear points.

Letδ(a, b) = 2. Then considering three pointsc, c^{0} and c^{00} on three different lines through a,
different from ab, shows that (a, b) ∈ O^{0}. Now let δ(a, b) = 8, and suppose (a, b) ∈ O^{0}. Let
m be any point at distance 4 from both a and b. Put a^{0} = proj_{m}a and b^{0} = proj_{m}b. Suppose
that the pointcmentioned in the property above lies at distance 5 from the linea^{0} (which is
allowed by Step 1 above). Then it is easy to see that c^{0} and c^{00} both have to lie at distance 5
fromb^{0}, which contradicts S_{c}^{0}_{,c}^{00}∩ {a, b}={a}. Similarly ifc lies at distance 5 from b^{0}.
Suppose from now on n >8. Let O^{0} be the subset of O of pairs (a, b) for which

(i) ∃!x_{1} ∈S_{a,b}:∀x^{0} ∈S_{a,b},|S_{x}_{1}_{,x}^{0} ∩ {a, b}|= 1,

(ii) there exist pointsc and c^{0} such that T_{a,b} ⊆Γ_{n−2}(c)∪Γ_{n−2}(c^{0}),
(iii) (x_{1}, v)∈O, with v an element of the set {a, b, c, c^{0}},

(iv) (a, c),(a, c^{0}),(b, c),(b, c^{0}),(c, c^{0})∈O,
(v) (c, y),(c^{0}, y),(x_{1}, y)6∈O, ∀y∈T_{a,b},
(vi) x_{1} =S_{a,c}∩S_{a,c}^{0} ∩S_{b,c}∩S_{b,c}^{0} ∩S_{c,c}^{0}.

We claim that O^{0} is the set of pairs of points at distance 2.

The claim is clear for two collinear points a and b. Indeed, let x1 be the unique point
on ab, different from a and b, and c and c^{0} points on two different lines (different from ab)
through the point x_{1} (Condition (v) in the definition of O^{0} does not hold if n = 8, which is
the reason we treated the octagons before).

So let δ(a, b) =n and suppose by way of contradiction that we have two points c and c^{0}
with the above properties. Let γ be a fixed n-path between a and b, and define m, a^{0} and b^{0}
as above in the case n = 8.

For an element x at distance j from m, 0 ≤j ≤n/2−3, such that proj_{m}a 6= proj_{m}x6=

proj_{m}b, we define the sets T_{x} in the same way as for the case s ≥ 3. We again first prove
thatTx6⊆Γn−2(c), for all sets Tx. Now, everything can be copied from the cases ≥3, except
when j = 0 or j = 2.

(j = 0) Suppose T_{m} ⊆ Γ_{n−2}(v), with v ∈ {c, c^{0}}. Then we know that δ(v, m) = n/2 and
we may assume proj_{m}v = proj_{m}a. This implies that δ(a, v) ≤n−2, so δ(a, v) = 4
(see Condition (iv)). Then x_{1} is the point of the path γ collinear with a (since S_{a,v}
contains only the element a1v in this case), so δ(b, x_{1}) = n−2, which contradicts
the fact that (x_{1}, b)∈O.

(j = 2) Note that this case is a problem only when n≡2 mod 4, so we may assume thatm
is a line. Suppose T_{L} ⊆Γ_{n−2}(v), for a line L concurrent withm, at distance n/2 + 2
from a and b, and for a point v ∈ {c, c^{0}}. Then δ(v, L) = δ(a, L) = n/2 + 2 and
proj_{L}a = proj_{L}v. We may again assume that v does not lie at distance n/2 from m.

By Step 1, there are essentially two possibilities for x_{1}.

First, suppose the point x1 lies at distance n/2 + 2 from m and at distance n/2 + 1
from a^{0}. Then there arises an n-path γ^{0} between a and x_{1} sharing the path [a, a^{0}]

withγ. Leta^{00}be the projection of x_{1} ontoa^{0}. Sincea^{00} is a line at distancen/2 from
both a and x_{1}, and v ∈S_{x}_{1}_{,a}, either the distance between v and a^{00} is n/2 (which is
not true), or the distance between v and a^{00} is n/2 + 2, which is again impossible.

Secondly,x_{1} cannot lie at distancen/2 fromm, since this would contradict condition
(v) (x_{1} would be collinear with a point of T_{a,b}).

Hence (3) is proved in this case.

Now we have to find an alternative argument for the last paragraph of the general case, since
we relied there on the fact that a line contains at least 4 points. We keep the same notation
of that paragraph. Now the only possibility (to rule out) that we have not considered yet
(because it does not occur in the general case) is the case that c and c^{0} both lie at distance
n−2 from different points u and u^{0} on L, δ(c, L) = δ(c^{0}, L) = n−1 and u and u^{0} different
from the projection w of a onto L.

Suppose first n > 10 (otherwise some of the notations introduced below don’t make
sense). Put L^{0} = proj_{w}a and l^{0} = proj_{L}0a. Suppose the unique point z on L^{0} at distance
n−2 from cis not l^{0}. Then consider a line K through z, different from L^{0} and from proj_{z}c.

Because c is at distance n from all the points of K, different from z (which are elements
of T_{a,b}) we can conclude that T_{K} ⊆ Γ_{n−2}(c^{0}), a contradiction to (3). So [c, L^{0}] contains l^{0}.
Define the element p as [l^{0}, m]∩[l^{0}, c] = [l^{0}, p]. Suppose p6=m and let j =δ(l^{0}, p). Consider
the element z^{0} on [c, p] at distancej+ 3 from p. Note that z^{0} is a line at distance n−3 from
both a and b. Since c is not at distance n−2 from any of the points of T_{z}^{0}, we conclude
that Tz^{0} ⊆ Γn−2(c^{0}), a contradiction to (3). If p = m, but if a^{0} 6= proj_{m}c 6= b^{0}, we obtain
a contradiction considering the line z^{0} at distance n/2−3 from m on [c, m] that does not
contain a^{0} orb^{0}). So the path [c, l^{0}] contains a^{0} orb^{0} (henceδ(c, m) = n/2 + 4). Suppose [c, l^{0}]
contains a^{0}. Consider now the element q defined by [m, c]∩[m, a] = [m, q]. Then we first
show that q coincides either with a^{0} (Case 2 below), or with the element a^{00} = proj_{a}0a (Case
1 below). Indeed, if not, then δ(a, c)< n, which implies that δ(a, c) = 4 (by Condition (iv))
and x1 = a1c. Since (b, x1) ∈ O, δ(b, x1) is then equal to n. So it would be possible to
find an element of T_{a,b} for which the projection onto ax_{1} is different from a and from x_{1},
a contradiction (such a point would be at distance n−2 from x_{1}, which would imply that
x1 6∈Sa,b). One checks that in the case n= 10, we end up with the same possibilities.

Case 1. Consider the element m^{0} ∈[a^{00}, c] that is at distance 2 from a^{00}. A point of S_{a,c} lies
at distance n/2 or n/2 + 2 from m^{0}. Because of the conditions, x_{1} ∈S_{a,c}. If x_{1} lies
at distance n/2 + 1 from a^{0}, then δ(x1, m^{0}) =n/2 + 4, a contradiction. If x1 lies at
distance n/2 + 1 from b^{0}, there arises a path of length n/2 + 6 between x_{1} and m^{0},
which is again a contradiction, since n >8. Note that x_{1} cannot lie at distance n/2
from m because (x1, y)6∈O for y∈Ta,b.

Case 2. Suppose x_{1} lies at distance n/2 + 1 from b^{0}. Let b_{0} be the projection of x_{1} onto
b^{0}. Then a point of S_{x}_{1}_{,b} lies at distance n/2 or n/2 + 2 from b_{0}. Because of the
conditions, c ∈ S_{x}_{1}_{,b}. But we have a path of length n/2 + 6 between c and b_{0}
(containing [c, a^{0}]), a contradiction since n 6= 8. Note that again, δ(x_{1}, m) 6= n/2.

So we know that x_{1} lies at distance n/2 + 1 from a^{0}. Suppose the projections of c
and x_{1} onto a^{0} are not equal (which only occurs if n ≡ 2 mod 4, since s = 2). Let
a_{0} = proj_{a}0x_{1}. Since c∈S_{x}_{1}_{,a}, the distance betweencanda_{0} is eithern/2 orn/2 + 2,

a contradiction (δ(c, a_{0}) = n/2 + 4). So the projection of c onto a^{0} is the element
a_{0}. Suppose proj_{a}_{0}c6= proj_{a}_{0}x_{1}. Since the distance betweencand a_{0} isn/2 + 2, and
c∈Sa,x1, the pointc has to lie at distance n/2 + 1 from either a^{0} or proj_{a}_{0}x1, which
is not true. So proj_{a}_{0}c= proj_{a}_{0}x_{1} :=h. Note that the projections of cand x_{1} onto
h are certainly different, since we know that the distance between c and x_{1} is either
n or 2, and the last choice would contradict the fact that a ∈ Sx1,c. Now consider
the projection m^{0} of c onto h. This is an element at distance n/2 from both c and
x_{1}. Now δ(b, m^{0}) = n/2 + 4, which contradicts the fact thatb ∈S_{c,x}_{1}.

This completes the case s= 2 and hence the casei=n−2.

2.6. Case i = n−1

We can obviously assume n ≥ 6. If n = 6 and s = t = 2, then an easy counting argument yields the result. Ifs= 2, then, since both sand tare infinite forn odd,n is even and hence t > 2. In this case, we dualize the arguments below (this is possible since i is odd). So we may assume throughout thats >2.

For two points a and b with δ(a, b) 6=n−1, let O_{a,b} be the set of pairs of points {c, c^{0}},
cand c^{0} different from a and from b, for which

T_{v,v}^{0} ⊆Γ_{n−1}(w)∪Γ_{n−1}(w^{0}),

whenever {a, b, c, c^{0}}={v, v^{0}, w, w^{0}}. For a pair {c, c^{0}} ∈O_{a,b}, we claim the following:

(i) If δ(a, b) = 2, then either c and c^{0} are different points on the line ab (distinct from a
and b), or, without loss of generality, c is a point onab and c^{0} ∈Γ_{3}(ab) with proj_{ab}c^{0} 6∈

{a, b, c}. Moreover, all the pairs (c, c^{0}) obtained in this way are elements of O_{a,b}.
(ii) If δ(a, b) = 4, then either c and c^{0} are collinear points on the lines am or bm (where

m=a1b) different from m, or cand c^{0} are points collinear with m, at distance 4 from
botha andb, and at distance 4 from each other. Again, all the pairs (c, c^{0}) obtained in
this way, are elements of O_{a,b}.

(iii) Let 4 < δ(a, b) = k < n−1 and put m = a1b. Then c and c^{0} are points at distance
k/2 from m, at distancek from both a and b, and at distancek from each other.

Ifδ(a, b) = 2, then an elementx ofT_{a,b}is either opposite the line ab, or lies at distance n−3
from a unique point on ab, different from a and from b. If δ(a, b) = 4, then an element x of
Ta,b either lies at distance n−1 or n−3 from m =a1b with proj_{m}a 6= proj_{m}x6= proj_{m}b or
lies at distance n−3 from a point x^{0} on am orbm, x^{0} 6∈ {a, b, m} with am6= proj_{x}0x6=bm.

It is now easy to see that the given possibilities forc and c^{0} in (i) and (ii) indeed satisfy the
claim for δ(a, b) = 2 and δ(a, b) = 4, respectively.

Letδ(a, b) = k ≤n−2 and again putm =a1b. Suppose {c, c^{0}} ∈O_{a,b}. For an element y
with δ(m, y) =j ≤n−k/2−2 and proj_{m}a 6= proj_{m}y6= proj_{m}b, we define the following set:

Ty ={x∈ P |δ(x, y) = (n±1)−j−k/2 and proj_{y}x6= proj_{y}m if δ(x, y)6=n}.

For an element y with δ(m, y) = n−k/2− 1 and proj_{m}a 6= proj_{m}y 6= proj_{m}b, we define
T_{y} as the set of elements at distance 2 from y, not incident with proj_{y}m. For an element y

with δ(m, y) = n−k/2 and proj_{m}a 6= proj_{m}y 6= proj_{m}b, we define T_{y} as the set of elements
incident with y, different from proj_{y}m. Note that T_{y} ⊆T_{a,b}.

First we make the following observation. Let y be an element for which the set Ty is
defined, and for which δ(m, y)≤n−k/2−2. Then there exists an element v ∈ {c, c^{0}} such
that T_{y} ⊆Γ_{n−1}(v) if and only if δ(v, y) = δ(a, y) and proj_{y}v = proj_{y}a or proj_{y}v = proj_{y}b.

Now we prove claims (i), (ii) and (iii) above by induction on the distance k between a and b. Let k≥2. In the sequel, we include the proof for the case k = 2 in the general case.

Suppose first there exists an element v ∈ {c, c^{0}} such that T_{m} ⊆ Γ_{n−1}(v). Then, by the
previous observation, δ(v, m) = δ(m, a) = k/2 and we may assume that proj_{m}v = proj_{m}a.

This implies that δ(a, v) ≤ k − 2. Put {c, c^{0}} = {v, v^{0}}. If k = 2, we obtain a = v, a
contradiction. If k = 4, then v is a point on the line am, v 6= m, and the only remaining
possibility, considering the induction hypothesis and the condition Ta,v ⊆Γn−1(b)∪Γn−1(v^{0})
is thatv^{0} is also a point onam, different from m. Ifk > 4, the position ofbcontradicts again
the fact thatT_{a,v} ⊆Γ_{n−1}(b)∪Γ_{n−1}(v^{0}) and the induction hypothesis. Indeed, the element at
distance δ(a, v)/2 from both a and v does not lie at distance δ(a, v)/2 from b. In this way,
we described all the possibilities for the points c and c^{0} in case there is a point v ∈ {c, c^{0}}
for which T_{m} ⊆ Γ_{n−1}(c). So from now on, we assume that there does not exist an element
v ∈ {c, c^{0}} such that Tm ⊆Γn−1(v).

Let l be any element incident with m, different from the projection of a or b onto m.

Suppose there exists a point v ∈ {c, c^{0}} such that T_{l} ⊆ Γ_{n−1}(v). Then δ(v, l) = δ(l, a) =
k/2 + 1 and we can assume that proj_{l}v = proj_{l}a = m. Since Tm 6⊆ Γn−1(v), we also know
that proj_{m}a6= proj_{m}v =:w6= proj_{m}b. Put {v, v^{0}}={c, c^{0}}.

Suppose firstk = 2. Thenv is a point on the lineab. We now show that the point v^{0} lies
at distance 2 or 4 fromv such that proj_{v}v^{0} =m. Indeed, suppose proj_{v}v^{0} 6=m orδ(v, v^{0}) =n.

If δ(v, v^{0}) 6=n, put γ^{0} = [v, v^{0}]. If δ(v, v^{0}) =n, let γ^{0} be an arbitrary n-path between v and
v^{0} not containing m. Let x be an element of T_{a,b} at distance n−3 from v such that either x
lies on γ^{0}, or [v, x] contains γ^{0}. Then x is an element ofT_{a,b} not at distance n−1 from v or
v^{0}, a contradiction. So we can assume that proj_{v}v^{0} =m. Suppose now 4 < δ(v, v^{0}). Let Σ be
an arbitrary apartment through v and v^{0}. Then the unique element of Σ at distance n−3
from v and belonging to T_{a,b}, does not lie at distance n−1 from v^{0}, a contradiction, so the
distance between v and v^{0} is 2 or 4. Suppose δ(v, v^{0}) = 4 and proj_{ab}v^{0} =b. Then we obtain
a contradiction (with the induction hypothesis) interchanging the roles of b and v. So v^{0} is a
point onab, orv^{0} is a point at distance 3 from abfor which the projection ontoabis different
froma,b orv, as claimed in (i).

Suppose now k 6= 2. Let w^{0} = proj_{w}v. Since the distance between v and any element
of T_{w}^{0} is less than or equal to n−3, we have that T_{w}^{0} ⊆ Γ_{n−1}(v^{0}), from which follows that
δ(v^{0}, w^{0}) = δ(w^{0}, a) = k/2 + 2 and proj_{w}0v^{0} = proj_{w}0a = w. Since T_{m} 6⊆ Γ_{n−1}(v^{0}), we either
have that v^{0} is a point at distance k/2 from m for which the projection onto m is different
from w and proj_{m}a 6= proj_{m}v^{0} 6= proj_{m}b (as required in (ii) and (iii)), or v^{0} is a point at
distance k/2 + 2 from m for which the projection onto m is w. In the latter case, let z be
the projection of v^{0} onto w (then δ(v, z) = δ(v^{0}, z) = k/2) and consider an element x at
distance n−1−k/2−2 from z such that proj_{z}v 6= proj_{z}x 6= proj_{z}v^{0}. Then x is an element
of T_{a,b} at distance n−3 from both v and v^{0}, a contradiction. In this way, we described all
the possibilities for the points c and c^{0} in case there is a point v ∈ {c, c^{0}} and an element l

as above for which T_{l} ⊆ Γ_{n−1}(c). So from now on, we assume that there does not exist an
element v ∈ {c, c^{0}} such that T_{l}⊆Γ_{n−1}(v), for any l as above.

We now prove that

(3) if yis an element for which the set T_{y} is defined, withδ(m, y)>1, then there does not
exist a point v ∈ {c, c^{0}} such that T_{y} ⊆Γ_{n−1}(v).

This is done by induction on the distance j between y and m.

So let by way of contradiction l be an element at distance j from m, j > 1, for which
the set T_{l} is defined and such that there exists an element v ∈ {c, c^{0}} with T_{l} ⊆ Γ_{n−1}(v).

Put {v, v^{0}} = {c, c^{0}}. Let first j < n −k/2−1. Then δ(v, l) = δ(l, a) = k/2 +j and
w := proj_{l}v = proj_{l}a but u := proj_{w}v 6= proj_{w}a. Let w^{0} = proj_{u}v. Note that the distance
between w^{0} and an element of T_{w}^{0} is (n ± 1)− k/2 −(j + 1), so an element of T_{w}^{0} lies
at distance at most n −3 from v. We conclude that Tw^{0} ⊆ Γn−1(v^{0}), from which follows
that δ(v^{0}, w^{0}) = δ(a, w^{0}) = k/2 +j + 1 or δ(v^{0}, w^{0}) = n− 3 (the latter is possible only if
j =n−k/2−2), and proj_{w}0v^{0} = proj_{w}0a=u. Let proj_{w}a=u^{0}. First supposeδ(v^{0}, w^{0})6=n−3.

From the assumptions, it follows that proj_{w}v^{0} 6=u^{0}. Depending on whether the projection of
v^{0} onto w is u or not, the distance between v^{0} and u^{0} is k/2 +j + 2 or k/2 +j. Note that
δ(v, u^{0}) = k/2 +j. Now consider an element x at distance (n−1)−(k/2 +j) from u^{0} such
that proj_{u}0x6=w, and such that x either lies on [u^{0}, b], or [u^{0}, x] contains [u^{0}, b]. Then x is an
element of T_{v,v}^{0} not contained in Γ_{n−1}(a)∪Γ_{n−1}(b), a contradiction. Ifδ(v^{0}, w^{0}) =n−3, then
we similarly obtain a contradiction. So T_{y} 6⊆ Γ_{n−1}(v), for any element y at distance j from
m.

Now let j = n −k/2−1. Note that T_{l} consists of all elements at distance 2 from l,
not incident with l^{0} = proj_{l}m. Then δ(v, l) = n−1 or δ(v, l) = n−3, and in both cases,
proj_{l}v = proj_{l}a. If δ(v, l) = n−1 (= δ(a, l)), we proceed as in the previous paragraph and
end up with a contradiction. So let δ(v, l) =n−3. First suppose that proj_{l}0v 6= proj_{l}0a=w.

Now consider an element w^{0} incident with w,l^{0} 6=w^{0} 6= proj_{w}a and w^{0} 6= proj_{w}b. ThenT_{w}^{0} ⊆
Γn−1(v), a contradiction sinceδ(m, w^{0}) =j −1. So proj_{l}0v =w. Let [u, m] = [v, m]∩[w, m]

and put u^{0} = proj_{u}v. Suppose first that proj_{m}a6=u^{0} 6= proj_{m}b and v 6=m (v =m can occur
only if k = 4). Then T_{u}^{0} ⊆ Γ_{n−1}(v^{0}). Indeed, if we put i= δ(u, l), then δ(v, u^{0}) = n−4−i
andδ(m, u^{0}) = n−k/2−i. So the distance betweenu^{0} and an element ofTu^{0} isi±1, and the
distance between v and an element of T_{u}^{0} is at most n−3. So T_{u}^{0} is contained in Γ_{n−1}(v^{0}),
which is a contradiction since δ(m, u^{0}) < j (indeed, i ≥ 2). Suppose finally u^{0} = proj_{m}a or
v =m. If k= 2, we end up with a pointv lying on ab(namelyv = proj_{m}l). But then, for an
arbitrary pointxonm, different froma,b andv, we have thatT_{x} ⊆Γ_{n−1}(v), in contradiction
with our assumptions. Ifk= 4, we end up with v =m, but then the position ofbcontradicts
the fact that Ta,v ⊆ Γn−1(b)∪Γn−1(v^{0}) and the (general) induction hypothesis. Finally, if
k >4, then δ(v, a)≤δ(v,proj_{m}a) +δ(a,proj_{m}a) =k−4. Now the position of b contradicts
again the fact that T_{a,v} ⊆Γ_{n−1}(b)∪Γ_{n−1}(v^{0}) and the (general) induction hypothesis.

Let finally j = n−k/2. Note that Tl consists of all elements incident with l, different
from the projection l^{0} of m onto l. Then δ(v, l^{0}) = n−1 or δ(v, l^{0}) = n−3. Note that, in
both cases, proj_{l}0v 6= proj_{l}0a. Indeed, proj_{l}0v = proj_{l}0a would imply that T_{l}^{0} ⊆ Γ_{n−1}(v), a
contradiction with our assumptions. Suppose firstδ(v, l^{0}) = n−3. Letl^{00}= proj_{l}0v. Since no
element incident with l^{00} is at distance n−1 from v, we have T_{l}^{00} ⊆Γ_{n−1}(v^{0}), which implies

thatδ(v^{0}, l^{0}) is eithern−3 orn−1 and proj_{l}0v^{0} 6= proj_{l}0a. Consider now the element on [a, l^{0}]
at distance 2 from l^{0}. This is an element of T_{v,v}^{0} which is at distance n−3 from botha and
b, a contradiction. Suppose now δ(v, l^{0}) = n−1. Let x be the element on [v, l^{0}] at distance
2 from l^{0}. Since xis the only element of T_{l}^{0} not at distance n−1 from v, this elementx lies
at distance n−1 from v^{0}. But then δ(v^{0}, l^{0}) is either n−1 or n−3. If proj_{l}0v^{0} = proj_{l}0a,
then Tl^{0} ⊆ Γn−1(v^{0}), a contradiction with our assumptions. If proj_{l}0v^{0} 6= proj_{l}0a, then again
the element on [a, l^{0}] at distance 2 froml^{0} is an element of T_{v,v}^{0} at distance n−3 from both
a and b, the final contradiction. This proves (3).

Suppose nowTy 6⊆Γn−1(v) for allv ∈ {c, c^{0}}and for any appropriate elementy. Consider
an elementlat distancen−k/2 frommsuch that the projection ofl ontomis different from
the projections ofaand bontom. Letube the projection ofaontol. SinceT_{l} 6⊆Γ_{n−1}(c) and
Tl 6⊆Γn−1(c^{0}), there is an elementxincident with l, different fromu, at distancen−1 fromc
but not fromc^{0}, and an elementy incident withl, different from u, at distance n−1 from c^{0}
but not from c. Soδ(x, c^{0}) =n−3 =δ(y, c) and proj_{x}c^{0} 6=l 6= proj_{y}c. But from this follows
that, for an arbitrary element l^{0} incident with u, l 6= l^{0} 6= proj_{u}a, we have Tl^{0} ⊆ Γn−1(c), a
contradiction. This proves the claims (i), (ii) and (iii).

For two points a, b, let C_{a,b} be the set containing a, b, and all points c for which there
exists a point c^{0} such that {c, c^{0}} ∈ Oa,b. Now let S be the set of pairs of points (a, b),
δ(a, b)6=n−1, for which there does not exist an element at distancen−1 from all the points
of C_{a,b}. We claim that S contains exactly the pairs of points (a, b) for which δ(a, b) = 2 or
δ(a, b) = 4.

First assumeδ(a, b) = 2. Let by way of contradiction w be an element at distance n−1
from all points ofC_{a,b}. Since all the points of the lineabare contained inC_{a,b},wlies opposite
ab. If v is an arbitrary point on ab, different from a and from b, then the element on [v, w]

that is collinear withv, is contained inC_{a,b}, but lies at distancen−3 fromw, a contradiction.

Suppose nowδ(a, b) = 4. Let by way of contradictionwbe an element at distancen−1 from
all points ofC_{a,b}. Then w lies at distance n−1 from all the points collinear with m=a1b,
which is not possible. Finally suppose 4< δ(a, b) =k 6=n−1. Let a^{0} be the element on the
path [a, b] at distance k/2−1 froma, and xan element at distance (n−1)−(k/2−1) from
w with proj_{a}0a 6= proj_{a}0w 6= proj_{a}0b. Then w lies at distance n−1 from all points of C_{a,b}.
Our claim is proved.

Now let S^{0} be the subset ofS containing all the pairs (a, b) with the property that there
exist points x and x^{0} belonging to C_{a,b} such that (x, x^{0}) 6∈ S. Then S^{0} contains exactly the
pairs of collinear points. Indeed, if δ(a, b) = 2, we can find pointsxand x^{0} inC_{a,b}at distance
6 from each other, while if δ(a, b) = 4, then C_{a,b} ⊆ Γ_{2}(m). This completes the proof of the
case i=n−1.

2.7. Case i = n/2

Let a and b be two points at distance k, and m an element at distance k/2 from both a
and b. Then it is easy to see that, if k 6= n, an arbitrary element x ∈ T_{a,b} lies at distance
n/2−k/2 from m such that proj_{m}a 6= proj_{m}c 6= proj_{m}b. Now we define the set S_{a,b} as the
set of points c, a 6=c6=b, for which T_{a,b} ⊆Γ_{n/2}(c). Suppose first i is odd, and s6= 2. Let S